Computability and Logic, 4th edition by George S. Boolos, John P. Burgess, Richard C. Jeffrey, PDF

By George S. Boolos, John P. Burgess, Richard C. Jeffrey,

ISBN-10: 0511078064

ISBN-13: 9780511078064

ISBN-10: 0521809754

ISBN-13: 9780521809757

Now in its fourth version, this booklet has turn into a vintage as a result of its accessibility to scholars with no mathematical heritage, and since it covers not just the staple subject matters of an intermediate good judgment path equivalent to Godel's Incompleteness Theorems, but in addition a number of non-compulsory issues from Turing's conception of computability to Ramsey's theorem. John Burgess has stronger the e-book through including a range of difficulties on the finish of every bankruptcy.

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Extra info for Computability and Logic, 4th edition

Sample text

Started scanning a stroke, it erases it, then goes to the halted state 2 when it scans the resulting blank, which means halting in a nonstandard final configuration. 1. THE HALTING PROBLEM 37 machines just described will be M1 , M2 , M3 , and we have f 1 = f 2 = f 3 = the empty function. We have now indicated an explicit enumeration of the Turing-computable functions of one argument, obtained by enumerating the machines that compute them. The fact that such an enumeration is possible shows, as we remarked at the outset, that there must exist Turing-uncomputable functions of a single argument.

F m , . . , we may conclude that the supposition is false. 1 Theorem. The diagonal function d is not Turing computable. According to Turing’s thesis, since d is not Turing computable, d cannot be effectively computable. Why not? After all, although no Turing machine computes the function d, we were able compute at least its first few values. For since, as we have noted, f 1 = f 2 = f 3 = the empty function we have d(1) = d(2) = d(3) = 1. And it may seem that we can actually compute d(n) for any positive integer n—if we don’t run out of time.

2 Theorem. The halting function h is not Turing computable. Proof: By way of background we need two special Turing machines. The first is a copying machine C, which works as follows. Given a tape containing a block of n strokes, and otherwise blank, if the machine is started scanning the leftmost stroke on the tape, it will eventually halt with the tape containing two blocks of n strokes separated by a blank, and otherwise blank, with the machine scanning the leftmost stroke on the tape. Thus if the machine is started with .

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